Mathematics Solutions Solutions for Class 8 Maths Chapter 15 Compound Interest are provided here with simple step-by-step explanations. These solutions for Compound Interest are extremely popular among Class 8 students for Maths Compound Interest Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 8 Maths Chapter 15 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 8 Maths are prepared by experts and are 100% accurate.

#### Question 1:

Find the amount and the compound interest.

 No. Principal (₹) Rate (p.c.p.a.) Duration (Years) 1 2000 5 2 2 5000 8 3 3 4000 7.5 2

(1) Here, P = ₹ 2000; R = 5 % ; N = 2 years

Hence, Amount = ₹ 2205 and Compound interest = ₹ 205.

(2) Here, P = ₹ 5000; R = 8 % ; N = 3 years

Hence, Amount = ₹ 6298.56 and Compound interest = ₹ 1298.56

(3) Here, P = ₹ 4000; R = 7.5 % ; N = 2 years

Hence, Amount = ₹ 4622.50 and Compound interest = ₹ 622.50

#### Question 2:

Sameerrao has taken a loan of ₹12500 at a rate of 12 p.c.p.a. for 3 years. If the interest is compounded annually then how many rupees should he pay to clear his loan?

Here, P = ₹ 12500; R = 12 % ; N = 3 years

Hence, he should pay an amount of ₹ 17561.60 to clear his loan.

#### Question 3:

To start a business Shalaka has taken a loan of ₹8000 at a rate of 10$\frac{1}{2}$ p.c.p.a. After two years how much compound interest will she have to pay?

Here, P = ₹ 8000; R = 10$\frac{1}{2}$ % ; N = 2 years

Hence, she will have to pay a Compound interest of ₹ 1768.20 after 2 years.

#### Question 1:

On the construction work of a flyover bridge there were 320 workers initially. The number of workers were increased by 25% every year. Find the number of workers after 2 years.

Here, P = Number of workers initially = 320
A = Number of workers after 2 years
R = Rate of increase of number of workers per year = 25 %
N = 2 years

Hence, the number of workers after 2 years is 500.

#### Question 2:

A shepherd has 200 sheep with him. Find the number of sheeps with him after 3 years if the increase in number of sheeps is 8% every year.

Here, P = Number of sheeps initially = 200
A = Number of sheeps after 3 years
R = Rate of increase of number of sheeps per year = 8 %
N = 3 years

Hence, the number of sheeps after 3 years is 252.

#### Question 3:

In a forest there are 40,000 trees. Find the expected number of trees after 3 years if the objective is to increase the number at the rate 5% per year.

Here, P = Number of trees initially = 40,000
A = Number of trees after 3 years
R = Rate of increase of number of trees per year = 5 %
N = 3 years

Hence, the expected number of trees after 3 years is 46,305.

#### Question 4:

The cost price of a machine is 2,50,000. If the rate of depreciation is 10% per year find the depreciation in price of the machine after two years.

Here, P = Cost price of the machine = 2,50,000
A = Cost price after 2 years
I = Depreciation in price after 2 years
R = Rate of depreciation = 10 %
N = 2 years

Also,
I = P − A
= 250000 − 202500
= 47500

Hence, the depreciation in price of the machine after two years is Rs 47,500.

#### Question 5:

Find the compound interest if the amount of a certain principal after two years is ₹4036.80 at the rate of 16 p.c.p.a.

Here, P = Principal
A = ₹ 4036.80
I = Compound Interest
R = 16 %
N = 2 years

$\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}⇒4036.80=\mathrm{P}{\left(1+\frac{16}{100}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒4036.80=\mathrm{P}{\left(1+\frac{4}{25}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒4036.80=\mathrm{P}{\left(\frac{29}{25}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{P}=\frac{4036.80×25×25}{29×29}\phantom{\rule{0ex}{0ex}}⇒\mathrm{P}=3000$

Also,
I = A − P
= 4036.80 − 3000
= 1036.80

Hence, the compound interest is ₹ 1036.80

#### Question 6:

A loan of ₹15000 was taken on compound interest. If the rate of compound interest s 12 p.c.p.a. find the amount to settle the loan after 3 years.

Here, P = Principal = ₹ 15000
A = Amount
R = 12 %
N = 3 years

Hence, the amount is ₹ 21073.92

#### Question 7:

A principal amounts to ₹13924 in 2 years by compound interest at 18 p.c.p.a. Find the principal.

Here, P = Principal
A = ₹ 13924
R = 18 %
N = 2 years

$\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}⇒13924=\mathrm{P}{\left(1+\frac{18}{100}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒13924=\mathrm{P}{\left(1+\frac{9}{50}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒13924=\mathrm{P}{\left(\frac{59}{50}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{P}=\frac{13924×50×50}{59×59}\phantom{\rule{0ex}{0ex}}⇒\mathrm{P}=10000$

Hence, the principal is ₹ 10000.

#### Question 8:

The population of a suburb is 16000. Find the rate of increase in the population if the population after two years is 17640.

Here, P = Population of a suburb = 16000
A = Population after two years =17640
R = R %
N = 2 years

$\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}⇒17640=16000{\left(1+\frac{\mathrm{R}}{100}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{17640}{16000}={\left(1+\frac{\mathrm{R}}{100}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{441}{400}={\left(1+\frac{\mathrm{R}}{100}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{21}{20}\right)}^{2}={\left(1+\frac{\mathrm{R}}{100}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒1+\frac{\mathrm{R}}{100}=\frac{21}{20}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{R}}{100}=\frac{21}{20}-1\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{R}}{100}=\frac{1}{20}\phantom{\rule{0ex}{0ex}}⇒\mathrm{R}=5$

Hence, the rate of increase in the population is 5 p.c.p.a.

#### Question 9:

In how many years ₹700 will amount to ₹847 at a compound interest rate of 10 p.c.p.a.

Here, P = Principal = ₹ 700
A = Amount = ₹ 847
R = 10 %
N = N years

$\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}⇒847=700{\left(1+\frac{10}{100}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}⇒\frac{847}{700}={\left(1+\frac{1}{10}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}⇒\frac{121}{100}={\left(\frac{11}{10}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{11}{10}\right)}^{2}={\left(\frac{11}{10}\right)}^{\mathrm{N}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{N}=2$

Hence, the number of years is 2 years.

#### Question 10:

Find the difference between simple interest and compound interest on ₹20000 at 8 p.c.p.a.

Disclaimer: In the question "Time" is not given. So the question is solved taking time as 2 years, because simple interest and compound interest will be same for one year.

Here, P = Principal = ₹ 20000
R = 8 %
N = 2 years

Hence, the difference between simple interest and compound interest is ₹ 128.

View NCERT Solutions for all chapters of Class 8